$f(t) = -3t^{2}+2t+2(h(t))$ $g(t) = 2t^{2}-6t+2-4(f(t))$ $h(x) = -2x$ $ h(f(3)) = {?} $
First, let's solve for the value of the inner function, $f(3)$ . Then we'll know what to plug into the outer function. $f(3) = -3(3^{2})+(2)(3)+2(h(3))$ To solve for the value of $f$ , we need to solve for the value of $h(3)$ $h(3) = (-2)(3)$ $h(3) = -6$ That means $f(3) = -3(3^{2})+(2)(3)+(2)(-6)$ $f(3) = -33$ Now we know that $f(3) = -33$ . Let's solve for $h(f(3))$ , which is $h(-33)$ $h(-33) = (-2)(-33)$ $h(-33) = 66$